expected value of a random palindrome

Question

If you choose a 6-digit palindrome at random, what is the expected value for the number?

All possible palindromes are equally likely to be chosen. Beginning number must be NONZERO, so numbers like 012321 are NOT allowed.

I'm not sure where to start. What values and the probabilities of the values should I look at?

Answer 1

Well, first of all, you need to count all possible outcomes.

In other words, you need to count all possible palindromes. All your palindromes are uniquely defined by their first three digits. You have $9$ possibilities for the first digit (can't be zero) and $10$ possibilities for the second and third.

Secondly, since the numbers are chosen with equal probabilities, you need to count their sum. The hint for this part is that $${abccba} = 100001\cdot a + 10010\cdot b + 1100\cdot c.$$

Can you take it from here?

Answer 2

The smallest palindrome is $100001$.

The largest palindrome is $999999$.

The second smallest palindrome is $101101$.

The second largest palindrome is $998899$.

Note that for every palindrome $X$, there is a palindrome $Y$ such that $X+Y=1100000$.

Since the palindromes are uniformly distributed, their expected value is $\frac{1100000}{2}=550000$.

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Explanation:

Assume that there are $n$ palindromes $d_1,\dots,d_n$.

Divide them into $\frac{n}{2}$ pairs, where the sum of each pair is $1100000$.

The sum of all the palindromes is therefore $\frac{n}{2}\cdot1100000=550000n$.

Since the palindromes are uniformly distributed, the probability of each palindrome is $\frac{1}{n}$.

So the expected value is $\frac{1}{n}\cdot{d_1}+\dots+\frac{1}{n}\cdot{d_n}=\frac{1}{n}\cdot(d_1+\dots+d_n)=\frac{1}{n}\cdot550000n=550000$.

Answer 3

There are 9 choices for the first digit, and 10 choices each for the second and third digit. There is an equal probability of each of these $900$ numbers.

So you want: $\sum\limits_{f=1}^9 \sum\limits_{s=0}^9\sum\limits_{t=0}^9 \left(\dfrac{100001\; f+10010\;s+1100\;t}{900}\right) \\ = \dfrac{100001\left(\sum\limits_{f=1}^9 f\right) \left(\sum\limits_{s=0}^9\sum\limits_{t=0}^9 1\right)+11110\left(\sum\limits_{f=1}^9\sum\limits_{k=0}^9 1\right) \left(\sum\limits_{h=0}^9 y\right)}{900}$

Can you complete?